Monday 13 November class

 

I apologize for the “Formula does not parse” and the failure to interpret latex errors below, which are occurring due to glitches in WordPress and I cannot figure out how to make them go away. I want to post this without further delay anyway.

Topics:

• two “distributed practice” review problems:

Determine whether the series converges or diverges, and find the sum if it converges: \displaystyle\sum_{n=1}^{\infty}\frac{4^{n}}{7^{n-1}}

Find the 4th degree Maclaurin polynomial for f(x) = \sqrt{x+1}

• New topic: Alternating series, the alternating series test for convergence, and estimating sums of alternating series.

Note: we are doing the topics in a different order than the order they are listed on the Course Outline (and in the textbook) because the Ratio and the Root tests are really tests for absolute convergence, and it seems better to introduce them only after we have studied absolute convergence.

 

Alternating series are simply series where consecutive terms have alternating signs. We have already seen examples of this. There are two very common ways to write alternating series, namely, if \{b_{n}\} is a positive sequence, we can write an alternating series as either

\displaystyle\sum_{n=1}^{\infty}(-1)^{n}b_{n} — this will make the first term have a negative sign and the second term have a positive sign, and so on

or we can write

\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}b_{n} — this will make the first term have a positive sign and the second term have a negative sign, and so on.

The second version is more common. The point is that when you see a series written in this kind of way, with a (-1) to the n, n+1, or even n-1 power included, you should recognize that it is an alternating series.

 

The alternating series test: For an alternating series, if the absolute values of the terms form a sequence which is decreasing and whose limit is 0, then the alternating series converges.

Note: there are two conditions that must be checked in order to use this theorem. Also, it cannot tell if the series diverges: it can only sometimes tell you that the series converges. If either of the two conditions fails, you must use some other test.

 

Examples:

The alternating harmonic series \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} converges, because the absolute values of the terms are

1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots

and this sequence is clearly decreasing and its limit is 0. So the alternating harmonic series converges. This does not tell us what it converges to: in fact, the sum is \ln(2) as we will show later.

(So even though the harmonic series diverges, the alternating harmonic series converges. Keep this in mind when we get to absolute convergence.)

 

Now consider the alternating series \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{|\sin(n)|}{n^{2}}. (We suspect this converges: do you see why?)

Can we use the alternating series test on this? The absolute values of the terms have the form $latex \frac{|\sin(n)|}{n^{2}}$, namely they form the sequence

These do not form a decreasing sequence, as we can see by looking at the graph:

Remember we are only interested in the points where the x-value is a natural number: but even for those (I have marked off the first few of them) they are not decreasing and it is not clear that even if we go far out in the graph they will ever form a decreasing sequence. So we cannot use the absolute value test for this alternating series: we will have to use some other test (and we will do so!)

 

Now consider the alternating series \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(n)}{n} Can we use the alternating series test?

The absolute values of the terms have the form $latex \frac{\ln(n)}{n} = 0, \frac{\ln(2)}{2}, \frac{\ln(3)}{3}, \dots$

We suspect that they form a decreasing sequence, at least eventually, because the denominator is growing faster than the numerator. By looking at the graph, we can see that they do start to decrease but not right away: they seem to start decreasing at some point after n=2.

(In fact, you can show by using the derivative of f(x)= \frac{\ln(x)}{x} that the maximum occurs at x=e, so the terms will in fact be decreasing starting with n=3. And you should do this! It is not good to put too much trust in graphs alone. I’ve marked that maximum point on the graph linked above.)

So the sequence of absolute values is decreasing from n=3 onward. It is always OK when determining convergence if we ignore the first few terms of a series: convergence is determined by what happens “in the long run”.

Also, \displaystyle\lim_{n\rightarrow\infty}\frac{\ln(n)}{n} = 0, so the second condition holds.

Therefore the alternating series \displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(n)}{n} converges. (Although we don’t know what it converges to!)

 

We have seen several times that we can know that a series converges without being able to know what it converges to. There is a theorem that tells us how good an alternating series’ partial sums are at estimating the sum of the series.

Estimating sums of convergent alternating series:

Suppose that \displaystyle\sum_{n=1}^{\infty}a_{n} is a convergent alternating series whose sum is L,:

L = \displaystyle\sum_{n=1}^{\infty}a_{n}

We want to estimate L by using the partial sum $S_{n}$.

Then the error we would make would be less than |a_{n+1}| in size: |S_{n}-L|<|a_{n+1}|

And we also know that the true sum L is between S_{n} and S_{n+1}:

S_{n} < L < S_{n+1}

Using this theorem and the alternating harmonic series, suppose we wanted to estimate the sum of the alternating harmonic series (which we have said, but not proved, is \ln(2)) to within 1/5. (In other words, we want our estimate to be no more than 1/5 = 0.2 away from the true value.)

Since 1/5 is the absolute value of the 5th term of the series, we will estimate by S_{4} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} \approx 0.583.

The theorem tells us that this estimate is no more than 0.2 off from the true value of the sum of the series (which is \ln(2), and also that the true value of the series is in between this number and S_{5} = \frac{7}{12} +\frac{1}{5} = \frac{47}{60} \approx 0.783.

In fact, \ln(2)\approx 0.693, so it is between  0.583 and 0.783, and the error we would make using 0.583 as an estimate for \ln(2) is |S_{4}-L| \approx |0.583 - 0.693| = 0.110 which is indeed less than 0.2.

 

Homework:

• Review the examples discussed in class (which are mostly detailed above).

• Also recommended: Paul’s Online Notes on Alternating series and the alternating series test.

Note: make sure that you understand when and how to use the alternating series test. We are not using the Ratio Test yet: please don’t try to bypass what we are doing right now, it will only harm you in the long run.

• You may also want to look at Paul’s Online notes for estimating alternating series: it is about halfway down in these notes.

And here is PatrickJMT on estimating alternating series

• Do at least the following problems from the WeBWorK on Alternating Series: #1-4 and #9. Note that in some problems you only get one attempt, so make it count! Also note that in problems 2 and 3 you are only to make use of the alternating series test, not some other test you may happen to know.

• Don’t forget that Test 3 is scheduled for Wednesday. Review materials are in this separate post.

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!